If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \ 0 & \omega\end{array} ight]$, then $A^{50}$ is equal to

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If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \ 0 & \omega\end{array}ight]$, then $A^{50}$ is equal to

Aicharya · Accepted Answer

$A=\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]$ Then, $A^2=\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]=\left[\begin{array}{cc}\omega^2 & 0 \ 0 & \omega^2\end{array}ight]$ and $A^3=A^2 \cdot A$ $ \begin{aligned} A^3 & =\left[\begin{array}{cc} \omega^2 & 0 \ 0 & \omega^2 \end{array}ight]\left[\begin{array}{cc} \omega & 0 \ 0 & \omega \end{array}ight] \ & =\left[\begin{array}{cc} \omega^3 & 0 \ 0 & \omega^3 \end{array}ight] \end{aligned} $ $A^3=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]\left[\because \omegaight.$ is cube root of unity $\left.\Rightarrow \omega^3=1ight]$ Now, $A^{50}=A^{48} \cdot A^2$ $ \begin{aligned} & =\left(A^3ight)^{16} \cdot A^2=(I)^{16} \cdot A^2 \ & =I A^2=A^2 \ & =\left[\begin{array}{cc} \omega^2 & 0 \ 0 & \omega^2 \end{array}ight]=\omega\left[\begin{array}{ll} \omega & 0 \ 0 & \omega \end{array}ight] \ A^{50} & =\omega A \end{aligned} $

If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \\ 0 &…

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